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A particle A of mass 'm' and charge 'q' is accelerated by a potential difference of 50 V . Another particle B of mass '4 m' and charge 'q' is accelerated by a potential difference of 2500 V . The ratio of de-Brogile wavelengths \frac{\lambda_{A}}{\lambda_{B}} is close to :

  • Option 1)

     4.47

  • Option 2)

     10.00

  • Option 3)

     0.07

  • Option 4)

     14.14

Answers (1)

best_answer

 

De - Broglie wavelength -

\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}

- wherein

h= plank's\: constant

m= mass \: of\: particle

v= speed \: of\: the \: particle

E= Kinetic \: energy \: of \: particle

\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mqv}}

Where V is the potential

\frac{\lambda _{A}}{\lambda _{B}}=\frac{\sqrt{2m_{B}q_{B}V_{B}}}{2m_{A}q_{A}V_{A}}=\sqrt{\frac{4m_{q}\times 2500}{m_{q}50}}=2\sqrt{50}

=14.14

 


Option 1)

 4.47

Option 2)

 10.00

Option 3)

 0.07

Option 4)

 14.14

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